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# Can some one pls help with this integration watch

1. ∫( (e^x)/((e^x)-(e^-x))dx

2. e^x/( e^x - e^(-x) ) = e^2x / ( e^2x - 1 )

So, the required integral is

INT e^2x / ( e^2x - 1 ) dx

You may be able to recognise that the numerator is half the derivitive of the denominator and then get the result straight away. If not, let u=e^2x, then x=(1/2)ln(u) and dx/du = 1/2u. Substitute these in

INT e^2x / ( e^2x - 1 ) dx = INT e^2x / ( e^2x - 1 ) (dx/du) du = INT u/(u-1) (1/2u) du = (1/2) INT 1/(u-1) du = (1/2)ln|u-1| + C = (1/2)ln|e^2x - 1| + C

I hope this helps.
3. ∫( (e^x)/((e^x)-(e^-x))dx
couldnt you rewrite this as the integral of e^x/cosh x then use parts?

what module is it from?
4. to me it looks like P3 (on edexcel at least), in which case hyperbolics aren't covered till P5
5. (Original post by rockindemon)
∫( (e^x)/((e^x)-(e^-x))dx
couldnt you rewrite this as the integral of e^x/cosh x then use parts?

what module is it from?
e^x/(e^x-e^(-x)) = e^x/(2sinh(x))

But also, this is P3 and I think my method above was simpler.
6. 1/2 ln|e^2x-1| + C = 1/2 (C + x + ln|e^x - e^-x|)

sorry, i just think that looks more aesthetic, especially as it leads on to a log of a hyperbolic sine, 1/2 (C + x + ln|sinh(x)|)

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Updated: April 19, 2004
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